Question

find a 2 digit no whose ten digit is 3 more than unit digits and no itself is 7 times sum of digit​

Answer 1

The unit digit of a two-digit number is 3

u = 3

and seven times the sum of the digits is the number itself.

7(t+u) = 10t+u

Substitute 3 for u:

7(t+3) = 10t+3

7t+21 = 10t+3

-3t = -18

t = 6

The tens digit is 6 and the units digit is 3, so the number is 63.

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Answer 2

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